Norm bound on exponential matrix with eigenvalue negative real part, proof

Clearly, if $A$ is diagonalizable then $J$ is diagonal and $\|e^{Jt}\|\le e^{-at}$ with $-a=\max{\text{Re}\lambda_i}$ (the slowest exponential).

If not then it is block diagonal and it is enough to bound each block. Let $J_k=\lambda I+N$ be a Jordan block of size $k$, $\lambda$ be the eigenvalue and $N$ is a matrix of all zeros except identities on the first superdiagonal. Then $$ e^{J_kt}=e^{(\lambda I+N)t}=[\text{$I$ and $N$ commute}]=e^{\lambda t}e^{Nt}= e^{\lambda t}\sum_{m=0}^k\frac{N^m}{m!}t^m. $$ The series for $e^{Nt}$ ends after $k$-th power since $N^k=0$. It gives for any small $\epsilon>0$ $$ \|e^{J_kt}\|\le e^{(\text{Re}\lambda+\epsilon)\cdot t}\left\|\sum_{m=0}^k\frac{N^m}{m!}t^me^{-\epsilon t}\right\|\le Ce^{(\text{Re}\lambda+\epsilon)\cdot t}. $$ Now pick $\epsilon>0$ small enough to get $\max\text{Re}\lambda_i+\epsilon=-\alpha<0$.